The Wheel of Fortune involves the repeated spinning of a wheel with 72 possible

Question

The Wheel of Fortune involves the repeated spinning of a wheel with 72 possible stopping points. We assume that each time the wheel is spun, any stopping point is equally likely. Exactly one stopping point on the wheel rewards a contestant with $1,000,000. Suppose the wheel is spun 24 times. Let X be the number of times that someone wins $1,000,000. Using the Poisson Approximation the Binomial, estimate the following probabilities: P(X =0), P(X = 1), P(X = 2). Count the number of distinct ways in which you can arrange the letters of the words: CATTERPILLAR, and ARUGULA.

Explanation / Answer

Exercise 1 . Probability of wheel stopping at particular point = 1/72

Number of times wheel is spun = 24 times.

so Expected number of times when wheel will stop at that particular point = 24 * (1/72) = 1/3

as we are doing Poisson approximation of binomial

Here = 1/3

so P(X=0) = e- ()x /x! = e-1/3(1/3)0 /0! = 0.7165

P ( X=1) = e-1/3(1/3)1 /1! = 0.2388

P(X=2) = e-1/3(1/3)2 /2! = 0.0398

Exercise 2

Number of distinct ways one can arrange word CATTERPILLAR

Total letters = 12 and T = 2 times, L = two times and A = 2 times and R = 2 times

so number of distinct ways it can be written = 12!/ 2! 2! 2! 2! = 29937600

NUmber of distinct ways one can arrange letters of word ARUGULA

here total letters = 7 and A = 2 times, U - 2 times and rest are one times

so number of distinct ways it can be written = 7!/ 2! 2! = 1260

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